# Full text of "Text Book Of Mechanical Engineering"

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```Examples in Tension -f Bending.

455

Apparently, stresses of 10, 7, and 3 tons are experienced with
the given loads, if the co-efficient O = 2 be used. But if the
bending formula be true without 0, stresses of 17^, i2-|, and 5
tons are involved. Now the elastic limit for wrought iron is from
12 to 15 tons, so the bending theory would appear to be insufficient,
and further that engineers are still wise in designing hooks and
such constructions by reference to the breaking load, on which
a factor of not less than 6 is adopted.

Example 31.—A longitudinal steel boiler stay, 20 ft. long and 2"
diameter, supports a flat area of 15 ins. sq., having on it a pressure of
120 Ibs. per sq. in. Find the greatest stress in the stay due to its own
weight and the steam pressure. (Hons. Mach. Constr. Exam. 1890.)

Weight of stay <w — 20 x 12 x 3*14 x '29 = 218*5 Ibs.
Steam pressure P =      15x15x120      = 27,000 Ibs.
^        w I       . TT d?        ,   _     14 wl      _       .,

=/>-—    and/o^TTwT = 8342 Ibs.

m ~~   8
ft a = 27,000
Total stress =/+/>

and

3'H

= 8598

' 8598-1-8342 = 16,940 Ibs. = 7*56 tons.

Example 32.—A piece of T iron consists of a web 4" deep and
f" thick, and a flange 2" wide and %' thick. Compare its strength
under longitudinal pull for the two cases (i) with line of action
through centre of web depth ; (2) with line of action passing through
centre of figure of the T. (Hons. Mach. Constr. Ex. 1888.)

See Fig. 414.

Find neutral axis by taking moments round A : k=i75". Draw
lines of limiting stress and find stress areas.

Z = area x arm = '6875 x 3*205 = 2*203
a = 3 sq. ins.    and r = 75"

Case (i)   /max =—H

3
W

1          Case (2)   /max =  — :

2-203

_
' Streng3hT(25 "

or as i : 2 roughly
•67

Combined Bending and Compression Stress-Action

is calculated by the same formula as for tension and bending, by
substituting^ in the direct stress.```